12x^2-88x+40=0

Solution for 12x^2-88x+40=0 equation:

12x^2-88x+40=0

a = 12; b = -88; c = +40;
Δ = b2-4ac
Δ = -882-4·12·40
Δ = 5824
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{5824}=\sqrt{64*91}=\sqrt{64}*\sqrt{91}=8\sqrt{91}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-88)-8\sqrt{91}}{2*12}=\frac{88-8\sqrt{91}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-88)+8\sqrt{91}}{2*12}=\frac{88+8\sqrt{91}}{24}
$